1. Integration by Substitution
Antidifferentiation of a Composite Function
Let f and g be functions such that f og and g’ are
continuous on an interval I. If F is an antiderivative
of f on I, then
Inside
Derivative
of Inside
Outside

2. Evaluate
Let u = x2 + 1
du = 2x dx
Let u = 5x
Evaluate
du = 5 dx

3. Multiplying and dividing by a constant
Let u = x2 + 1
du = 2x dx
Let u = 2x - 1
du = 2dx

4. u = 3x - 1 u = x2 + x u = x3 - 2 u = 1 – 2x2 u = cos x
Substitution and the General Power Rule
What would you let u = in the following examples?
u = 3x - 1
u = x2 + x
u = x3 - 2
u = 1 – 2x2
u = cos x

5. A differential equation, a point, and slope field are given. Sketch
the solution of the equation that passes through the given point.
Use integration to find the particular solution of the differential
equation.
Day 1 stop (1-41 odd)

6. u = x3
du = 3x2 dx

7. Let u = sin 3x
du = 3cos 3x dx
rewritten as
Day 2 stop (43-56 all, odd)

8. Let u = 2x - 1
u + 1 = 2x
du = 2dx

9. Evaluate
u = x2 + 1
du = 2x dx
Note that there are no upper and lower
limits of integration.
We must determine new upper and
lower limits by substituting the old
ones in for x in u = x2 + 1 2
Or, we could use the old limits if we
substitute x2 + 1 back in. 1

10. 2u du = 2 dx u du = dx What limits are we going to use? 3 1
See area comparisons when using different upper and lower
limits. Page 302

11. Integration of Even and Odd Functions
If f is an even function, then
Ex.
Odd or Even?
If f is an odd function, then